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In an LR circuit, the current reaches three-fourths of its final steady value after 4 s. What is the time constant of the circuit?
- ln 2 s
- 2/ln 2 s
- 3/ln 2 s
- 4/ln 2 s
Correct answer: 2/ln 2 s
Solution
i = i0(1 - e^(-t/tau)); 3/4 = 1 - e^(-4/tau) -> e^(-4/tau) = 1/4 -> 4/tau = ln4 = 2 ln2 -> tau = 2/ln2 s.
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